Xét \(v=v_0+at\Rightarrow t=\dfrac{v-v_0}{a}\)
Thay t vào công thức \(s=v_0t+\dfrac{1}{2}at^2\) ta được :
\(s=v_0\cdot\dfrac{v-v_0}{a}+\dfrac{1}{2}a\cdot\left(\dfrac{v-v_0}{a}\right)^2\)
\(=\dfrac{v_0\cdot v-v_0^2}{a}+\dfrac{v^2-2v\cdot v_0+v_0^2}{2a}=\dfrac{v^2-v^2_0}{2a}\)(đpcm)